如何仅推送在我的数组中找不到的元素

Question

所以问题是要求我从单词中删除标点符号，我知道如何交叉引用 tho 数组以检查我的数组中是否存在要检查的元素，但我如何只推送不在标点符号数组中的值？

function removePunctuation(word){
  var punctuation = [";", "!", ".", "?", ",", "-"];
  var chars = word.split("");
  var puncRemoved = [];
  for(var i = 0; i < chars.length;i++){
    for(var j = 0; j < punctuation.length;j++) {
      if(punctuation[j].indexOf(chars[i]) !== 0) {
        puncRemoved.push(i)
      }
    }
  }
  return puncRemoved;
}

Answer 1

word.replace(/[;!.?,-]/g, '');

您可能会发现这个非常有趣的:D

Answer 2

下面是一个基于您的代码的解决方案：

function removePunctuation(word){
  var punctuation = [";", "!", ".", "?", ",", "-"];
  var chars = word.split("");
  var puncRemoved = [];
  for (var i = 0; i < chars.length; i++) {
    // Push only chars which are not in punctuation array
    if (punctuation.indexOf(chars[i]) === -1) {
        puncRemoved.push(chars[i]);
    }
  }
  // Return string instead of array
  return puncRemoved.join('');
}

实现此目的的另一种方法是：

function removePunctuation(word){
  var punctuation = [";", "!", ".", "?", ",", "-"];
  // Iterate and remove punctuations from the given string using split().join() method
  punctuation.forEach(function (p) {
    word = word.split(p).join('');
  });
  return word;
}

或者，正如另一个答案中所建议的：

function removePunctuation(word){
    return word.replace(/[;!.?,-]/g, '');
}

Answer 3

当数组中找不到值时，您需要推送该值，因此：

punctuation.indexOf(chars[i]) == -1

但是正则表达式似乎简单得多。

编辑

为了清楚起见，您需要遍历字符，并且仅在它们未出现在标点符号数组中时才推送它们：

function removePunctuation(word){
  var punctuation = [";", "!", ".", "?", ",", "-"];
  var chars = word.split("");
  var puncRemoved = [];
  for(var i = 0; i < chars.length;i++){
    if(punctuation.indexOf(chars[i]) == -1) {
      puncRemoved.push(chars[i])
    }
  }
  return puncRemoved;
}
var s = 'Hi! there. Are? you; ha-ppy?';
console.log(removePunctuation(s).join(''))